Answer
given,
number of turns, N = 400 turns
mean radius, R = 6 cm
Current, I = 0.25 A
relative permeability, K_m = 80
a) Magnetic field in the core
[tex]B =\dfrac{K_m\mu_o N I}{2\pi R}[/tex]
[tex]B =\dfrac{80\times 4\pi \times 10^{7}\times 400 \times 0.25}{2\pi\times 0.06}[/tex]
B = 2.67 x 10⁻² T
b) the amount of magnetic field due to atomic current
[tex]B = \dfrac{K_m\mu_o N I}{2\pi R} - \dfrac{\mu_o N I}{2\pi R}[/tex]
[tex]B = \dfrac{(K_m-1)\mu_o N I}{2\pi R}[/tex]
[tex]B = \dfrac{(80-1)\times 4\pi \times 10^{7}\times 400 \times 0.25}{2\pi\times 0.06}[/tex]
B = 2.63 x 10⁻² T
